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Explicitly describing the singular locus from a finite set of polynomials

  1. Feb 17, 2012 #1
    When explicitly given a set of polynomial equations, I am interested in describing its singular locus.

    I read this from several sources that a point is singular if the rank of a Jacobian at a singular point must be any number less than its maximal possible number. Or is it the locus where all the 2x2 minors of a Jacobian vanish? How are the two related?

    For example, consider the scheme defined by these three equations: $r_{12}s_{21}+x_1y_1, (r_{22}-r_{11})s_{21}+x_2 y_1, -s_{21}r_{12}+x_2 y_2$ sitting in $\mathbb{C}^{10}$.

    Then the Jacobian is
    \[ J =
    \left[
    \begin{array}{cccccccccc}
    0 & s_{21} & 0 & 0 &r_{12} & 0& y_1& 0 &x_1 &0 \\
    -s_{21}& 0 &s_{21} &0 &r_{22}-r_{11} &0 &0 &y_1 &x_2 &0 \\
    0 & -s_{21} & 0& 0& -r_{12}&0 & 0& y_2& 0& x_2\\
    \end{array}
    \right].
    \]

    Denote the scheme defined by the three equations as X. If X were equidimensional, then is the singular locus defined by all the (10-dim X) = 3-minors of the Jacobian matrix?

    Or is the following procedure correct? First set all the entries in the first row equal to 0 and that is the singular locus on one of the irreducible components (assuming each equation is irreducible). Then set all the entries in the second row equal to 0, which will give the singular locus on another irreducible component. Do the same for the last row.

    Then set all the entries in any 2 rows equal to zero to obtain the singular locus where two hypersurfaces intersect.

    Finally, set all the entries in all the rows equal to zero to obtain the locus where all three hypersurfaces intersect.

    Isn't this second procedure (setting various rows equal to zero) more efficient than looking at all the 3x3 minors?

    Here is the locus defined by the set of all 3x3 minors and it seems quite messy:
    s21^2 y1, s21^2 y2, s21^2 x1, s21^2 x2, r12 s21 y1, r12 s21 y2, r12 s21 x1, r12 s21 x2, s21 y1 y2, s21 x2 y1, -s21 x1 y2, s21 x1 x2, s21^2 y1, s21^2 y2, s21^2 x1, s21^2 x2, (r22 - r11) s21 y1, (r22 - r11) s21 y2, (r22 - r11) s21 x1, (r22 - r11) s21 x2, -s21 y1^2, -s21 x2 y1, s21 (x1 y1 - x2 y2), s21 x2 y1, s21 x2^2, -r12 s21 y1, -r12 s21 y2, -r12 s21 x1, -r12 s21 x2, -s21 y1 y2, 0, -s21 x2 y1, s21 x1 y2, 0, -s21 x1 x2, -r12 y1^2 + (r11 - r22) y1 y2, -r12 x2 y1, (r11 - r22) x2 y1, r12 (x1 y1 - x2 y2) + (r22 - r11) x1 y2, r12 x2 y1, (r11 - r22) x1 x2 + r12 x2^2, -x2 y1 y2, x2 y1^2, x2^2 y1, -x1 x2 y1

    In a more general setting, it will look quite messy and I personally feel a lot of intuition is lost when analyzing all these equations.

    For this entire argument and for the sake of simplicity, let us just assume that X is a complete intersection.


    By the way, what typesetting does Physics Forum use? Obviously, it isn't LaTeX. Is it HTML?
     
  2. jcsd
  3. Feb 18, 2012 #2
    The singular locus is where the rank of the jacobian is not maximal. I cant see why your other "procedure" would give the same result here.

    Then you give us this messy set of 3x3-minors. If you really mean to treat all your x,y,s,r, with multiple indices, as variables, without mentioning how this naturally appears, it would be a lot easier for both you and the reader, if you simplifed your notation, and instead gave the equivalent equations:

    ab+de = ab+fg = ac+df = 0

    Here we immediately see that the intersection contains four linears subspaces (a=0). Then eliminating one of the first equations, you may find that the remaining locus is an irreducible quartic (edit: or the union of a cubic and a linear space) with a linear singular locus. Tell us how that works out.
     
    Last edited: Feb 18, 2012
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